How do you integrate #int e^x/[(e^x-2)(e^(2x)+1)]dx# using partial fractions?

Question

17759 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-11-10T14:04:41

#1/5(lnabs(e^x-2)-1/2lnabs(e^(2x)+1)-2 arctan e^x)+c#

Substitution with #e^x=y# then partial fractions

#x=lny\ \ , \ \ dx=1/ydy#

#int1/((y-2)(y^2+1))dy#

Search #A,B,C# so that

#1/((y-2)(y^2+1))=A/(y-2)+(By+C)/(y^2+1)#
#1/((y-2)(y^2+1))=((A+B)y^2+(C-2B)y+A-2C)/((y-2)(y^2+1))#

So #A=-B#, #C=2B# and #A-2C=-B-4B=1#

Then #B=-1/5,A=1/5, C=-2/5#

#int1/((y-2)(y^2+1))dy=#
#=1/5(int1/(y-2)dy-1/2int(2y)/(y^2+1)dy-2int1/(y^2+1)dy)=#
#=1/5(lnabs(y-2)-1/2lnabs(y^2+1)-2 arctan y)+c#

Then substitute #y# with #e^x#